The equation #a^3+b^3+c^3=2008# has a solution in which #a,b,and c# are distinct even positive integers. find #a+b+c#?

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Answer 1 · 2018-06-18T09:20:08

The answer is #=22#

The equation is

#a^3+b^3+c^3=2008#

Since #a,b,c in NN# and are even

Therefore,

#a=2p#

#b=2q#

#c=2r#

Therefore,

#(2p)^3+(2q)^3+(2r)^3=2008#

#=>#, #8p^3+8q^3+8r^3=2008#

#=>#, #p^3+q^3+r^3=2008/8=251#

#=>#, #p^3+q^3+r^3=251=6.3^3#

Therefore,

#p#, #q# and #r# are #<=6#

Let #r=6#

Then

#p^3+q^3=251-6^3=35#

#p^3+q^3=3.27^3#

Therefore,

#p# and #q# are #<=3#

Let #q=3#

#p^3=35-3^3=35-27=8#

#=>#, #p=2#

Finally

#{(a=4),(b=6),(q=12):}#

#=>#, #a+b+c=4+6+12=22#